The Designer's Guide Community Forum https://designers-guide.org/forum/YaBB.pl Simulators >> Circuit Simulators >> op amp simulation question https://designers-guide.org/forum/YaBB.pl?num=1240846163 Message started by liletian on Apr 27th, 2009, 8:29am

Title: op amp simulation question Post by liletian on Apr 27th, 2009, 8:29am Hi Guys I simulated a op amp feedback circuit as attached, it is wired the output has DC offset of 0.004 voltage, can anyone explain it? The input is a Vsin of 2Ghz without dc offset in the negative input, the positive input is connect to ground, I will expect there has no output DC offset. Can anyone explain it? please see the attachment. The op amp is an ideal op amp with gain of 1e10. Thank you

Title: Re: op amp simulation question Post by LeeX on Apr 27th, 2009, 9:19am Is this op-amp ideal or is it some circuit you have designed? Also, for the configuration shown in the fig, what kind of task you expect the circuit to perform?

Title: Re: op amp simulation question Post by liletian on Apr 27th, 2009, 9:37am LeeX wrote on Apr 27th, 2009, 9:19am: Is this op-amp ideal or is it some circuit you have designed? Also, for the configuration shown in the fig, what kind of task you expect the circuit to perform? The Op-amp is ideal, it is mathmatic equation( using verilogA), I would expect this circuit to work as an amplifier with a gain. I would not like to see a DC offset in there. Very weird.  Thanks

Title: Re: op amp simulation question Post by raja.cedt on Apr 27th, 2009, 9:58am hi liletian,               i didn't understand what's your plan with this circuit?As you said if you want amplifier why did you kept that cap in the feedback and even you don't have any dc source in the circuit,how would you expect dc offset. Thanks, Rajasekhar.

Title: Re: op amp simulation question Post by liletian on Apr 27th, 2009, 10:17am raja.cedt wrote on Apr 27th, 2009, 9:58am: hi liletian,               i didn't understand what's your plan with this circuit?As you said if you want amplifier why did you kept that cap in the feedback and even you don't have any dc source in the circuit,how would you expect dc offset. Thanks, Rajasekhar. I do not expect DC offset, but my simulation gives me DC offset, that's why I am so confusing. :-[

Title: Re: op amp simulation question Post by buddypoor on Apr 27th, 2009, 10:20am Hi  liletian, as already mentioned before: what is the purpose of the feedback capacitor ? It destroys your dc operating point (no dc feedback). The circuit cannot work.

Title: Re: op amp simulation question Post by liletian on Apr 27th, 2009, 10:24am buddypoor wrote on Apr 27th, 2009, 10:20am: Hi  liletian, as already mentioned before: what is the purpose of the feedback capacitor ? It destroys your dc operating point (no dc feedback). The circuit cannot work. It can work, why it can not work, the signal is a RF signal. The circuit is for charge pump purpose.

Title: Re: op amp simulation question Post by salty on Apr 27th, 2009, 10:46am My guess is numerical resolution.  The only reason this ckt "works" is because everything is ideal.  If you see 4mV on the output, that would imply 400fV on the input which is about 2^-41.  I think you are just seeing numerical issues.  If you have very much DC at all on the input of the amplifier with 10^10 gain, the amp will rail.  You have no DC path around the amplifier.

Title: Re: op amp simulation question Post by buddypoor on Apr 27th, 2009, 1:33pm liletian wrote on Apr 27th, 2009, 10:24am: It can work, why it can not work, the signal is a RF signal. The circuit is for charge pump purpose. Please, excuse me. But I think its absolutely silly to use an artificial amplifier with a gain of 1E10 without a stable bias point and to worry about an offset of 4 mvolts. Simulation programs are a tool to simulate real circuits which cannot be analyzed with calculations by hand.

Title: Re: op amp simulation question Post by liletian on Apr 27th, 2009, 2:28pm buddypoor wrote on Apr 27th, 2009, 1:33pm: liletian wrote on Apr 27th, 2009, 10:24am: It can work, why it can not work, the signal is a RF signal. The circuit is for charge pump purpose. Please, excuse me. But I think its absolutely silly to use an artificial amplifier with a gain of 1E10 without a stable bias point and to worry about an offset of 4 mvolts. Simulation programs are a tool to simulate real circuits which cannot be analyzed with calculations by hand. for my defense, actually for a gain of 500, there still has 4 mV DC offset.

Title: Re: op amp simulation question Post by subgold on Apr 27th, 2009, 3:25pm liletian wrote on Apr 27th, 2009, 2:28pm: buddypoor wrote on Apr 27th, 2009, 1:33pm: liletian wrote on Apr 27th, 2009, 10:24am: It can work, why it can not work, the signal is a RF signal. The circuit is for charge pump purpose. Please, excuse me. But I think its absolutely silly to use an artificial amplifier with a gain of 1E10 without a stable bias point and to worry about an offset of 4 mvolts. Simulation programs are a tool to simulate real circuits which cannot be analyzed with calculations by hand. for my defense, actually for a gain of 500, there still has 4 mV DC offset. could you pls post your verilogA code?

Title: Re: op amp simulation question Post by boe on Apr 28th, 2009, 6:09am liletian wrote on Apr 27th, 2009, 2:28pm: ... for my defense, actually for a gain of 500, there still has 4 mV DC offset. BTW, what is your output voltage? If the Gain is actually 500 [at DC] and you want 2 V at the output, you need 4 mV at the input. B.O.E.

Title: Re: op amp simulation question Post by LeeX on Apr 28th, 2009, 11:02am I think I may have found how you get the 4 mV "offset". First of all, I do not think this configuration can work for any practical design, as no DC feedback is present. However, for the ideal op-amp, I assume it works properly and by "offset" you refer to the difference between zero cross points for the input and output. The transfer function between input and output is Vout/Vin=1+1/(jw/(1/RC)) where w is the angular frequency, in your case it is 2G*2*Pi, R=10K and C=200fF, according to your circuit. Thus Vout/Vin=1-j0.03978 So Vout is actually shifted in phase and amplified in magnitute, comparing to Vin, the phase shift is -0.03978 so if Vin = 0.1*Sin(wt) Vout is approximately Vout = 0.1*Sin(wt-0.03978) let t=0, Vin = 0 Vout = -0.00397≈4 mV Actually, by definition the word offset should refer to the voltage difference between the op-amp positive and the negative input when the output is zero (common mode). If my guess is right, you are measuring the voltage difference between the output node and input node, non of them are the opamp input.

Title: Re: op amp simulation question Post by liletian on Apr 28th, 2009, 1:35pm  HI  Thanks for answering, you might be right, I will think about it more. I do know there is no DC feedback, but it seems is necessary for a charge pump of PLL, not sure about it yet. Can you comment more on the DC feedback and why it should not working for any pratical circuits?  Thanks a lot LeeX wrote on Apr 28th, 2009, 11:02am: I think I may have found how you get the 4 mV "offset". First of all, I do not think this configuration can work for any practical design, as no DC feedback is present. However, for the ideal op-amp, I assume it works properly and by "offset" you refer to the difference between zero cross points for the input and output. The transfer function between input and output is Vout/Vin=1+1/(jw/(1/RC)) where w is the angular frequency, in your case it is 2G*2*Pi, R=10K and C=200fF, according to your circuit. Thus Vout/Vin=1-j0.03978 So Vout is actually shifted in phase and amplified in magnitute, comparing to Vin, the phase shift is -0.03978 so if Vin = 0.1*Sin(wt) Vout is approximately Vout = 0.1*Sin(wt-0.03978) let t=0, Vin = 0 Vout = -0.00397≈4 mV Actually, by definition the word offset should refer to the voltage difference between the op-amp positive and the negative input when the output is zero (common mode). If my guess is right, you are measuring the voltage difference between the output node and input node, non of them are the opamp input.

Title: Re: op amp simulation question Post by buddypoor on Apr 28th, 2009, 11:28pm liletian wrote on Apr 28th, 2009, 1:35pm:  HI  Thanks for answering, you might be right, I will think about it more. I do know there is no DC feedback, but it seems is necessary for a charge pump of PLL, not sure about it yet. Can you comment more on the DC feedback and why it should not working for any pratical circuits?  Thanks a lot [/quote] Each opamp needs dc feedback to fix the bias point. Without dc feedback the offset drives the output in saturation.

Title: Re: op amp simulation question Post by liletian on Apr 29th, 2009, 7:09am buddypoor wrote on Apr 28th, 2009, 11:28pm: liletian wrote on Apr 28th, 2009, 1:35pm:  HI  Thanks for answering, you might be right, I will think about it more. I do know there is no DC feedback, but it seems is necessary for a charge pump of PLL, not sure about it yet. Can you comment more on the DC feedback and why it should not working for any pratical circuits?  Thanks a lot Each opamp needs dc feedback to fix the bias point. Without dc feedback the offset drives the output in saturation.[/quote] I knew it, but in this application, I can not add a DC feedthrough, any suggestions?  Thanks

The Designer's Guide Community Forum » Powered by YaBB 2.2.2! YaBB © 2000-2008. All Rights Reserved.